Question

Because the cylinders in Ex. 8.4 are left rotating (at angular velocities w_a and w_b, say), there is actually a residual magnetic field, and hence angular momentum in the fields, even after the current in the solenoid has been extinguished. If the cylinders are heavy, this correction will be negligible, but it is interesting to do the problem without making that assumption.

(a) Calculate (in terms of w_a and w_b ) the final angular momentum in the fields. [Define $\mathit{\omega }\mathbf{=}\mathit{\omega }\hat{\mathbf{z}}$, sow_a and w_b could be positive or negative.]

(b) As the cylinders begin to rotate, their changing magnetic field induces an extra azimuthal electric field, which, in turn, will make an additional contribution to the torques. Find the resulting extra angular momentum, and compare it with your result in (a).

Short answer

(a) The final angular momentum in the fields is $L=\frac{{\mu }_0{\omega }_b{Q}^2\left(b^2-a^2\right)}{4\pi l}\hat{z}$.

(b) The resulting extra angular momentum is $L_{\text{tot}}=-\frac{{\mu }_0{Q}^2{\omega }_b}{4\pi l}\left(b^2-a^2\right)\hat{z}$. The reduction in the final angular momentum in part (b) is equal to the residual angular momentum in part (a).

Step-by-step solution

Expression for the final angular momentum in the fields:

Write the expression for the final angular momentum in the fields.

$\mathit{L}\mathbf{=}\mathbf{\int }\mathit{l}\mathit{d}\mathit{\tau }$ …… (1)

Here, l is the angular momentum density and $\int d\tau$is the cross-sectional area.

Determine the angular momentum density:

(a)

Write the expression for the angular momentum density.

I = r x g …… (2)

Here, g is the momentum density.

Write the expression for the momentum density.

$g={\epsilon }_0\left(E\times B\right)$ …… (3)

Here, E is the electric field, and B is the magnetic field produced by the solenoid at radius b.

Write the expression for the electric field.

$$\begin{gathered} E=\frac{1}{2\pi {\epsilon }_0}\frac{\lambda }{s}\hat{s} \\ E=\frac{Q}{2\pi {\epsilon }_{0}ls}\hat{s} \end{gathered}$$

Write the expression for the magnetic field produced by the solenoid at radius b.

$B={\mu }_{0}K\hat{z}$

Substitute the standard formula for K.

localid="1653482230148" $B={\mu }_0\left({\sigma }_b{\omega }_{b}b\right)\overbrace{z}$

Substitute all the known values in equation (3).

localid="1653482236658" $g={\epsilon }_0\left(\frac{Q}{2{\mathrm{\pi \epsilon }}_0/\mathrm{s}}\overbrace{s}\times \frac{{\mu }_0{\omega }_{b}Q}{2\mathrm{\pi I}}\overbrace{z}\right)$

localid="1653482242686" $g={\epsilon }_0\left(\frac{Q}{2{\mathrm{\pi \epsilon }}_{0}I\mathrm{s}}\right)\left(-\frac{{\mu }_0{\omega }_{b}Q}{2\mathrm{\pi I}}\right)\left(\overbrace{s}\times \overbrace{z}\right)$

localid="1653482248181" $g=\frac{{\mu }_0{\omega }_b{Q}^2}{4{\mathrm{\pi }}^2{\mathrm{I}}^2\mathrm{s}}\overbrace{\varnothing }$

Substitute the value of g in equation (2).

localid="1653482256169" $I=r\times \left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\mathrm{\pi }}^2{\mathrm{I}}^2\mathrm{s}}\overbrace{\varnothing }\right)$

localid="1653482262657" $I=\left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\mathrm{\pi }}^2{\mathrm{I}}^2\mathrm{s}}\right)\left(r\times \overbrace{\varnothing }\right)$

localid="1653482269792" $I=\left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\mathrm{\pi }}^2{\mathrm{I}}^2}\right)\overbrace{z}$

Determine the final angular momentum in the fields:

Substitute the known values in equation (1).

$L=\left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\mathrm{\pi }}^2{\mathrm{I}}^2}\right)\overbrace{Z}\int d\zeta$

localid="1653482294398" $$\begin{gathered} L=\left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\pi }^2{l}^2}\right)\hat{z}\int d\tau \\ L=\left(\frac{{\mu }_0{\omega }_b{Q}^2}{4{\pi }^2{l}^2}\right)\pi \left(b^2-a^2\right)l\hat{z} \\ L=\frac{{\mu }_0{\omega }_b{Q}^2\left(b^2-a^2\right)}{4\pi l}\hat{z} \end{gathered}$$

Therefore, the final angular momentum in the fields is localid="1653482299337" $L=\frac{{\mu }_0{\omega }_b{Q}^2\left(b^2-a^2\right)}{4\pi l}\hat{z}$.

Determine the extra electric field induced by the changing magnetic field and the extra angular momentum:

(b)

Write the expression for the extra electric field induced by the changing in the magnetic field due to the rotating shell.

$$\begin{gathered} \boxed{\int }E·dI=-\frac{d\varnothing }{dt} \\ E\left(2\mathrm{\pi s}\right)=-\frac{d\varnothing }{dt} \\ E=-\frac{1}{2\mathrm{\pi s}}\frac{d\varnothing }{dt} \end{gathered}$$

…… (4)

Write the expression for the electric flux .

$$\begin{gathered} \varnothing =\frac{{\mu }_{0}Q}{2\mathrm{\pi I}}\left({\omega }_a-{\omega }_b\right){\mathrm{\pi a}}^2-\frac{{\mathrm{\mu }}_0{\mathrm{Q\omega }}_{\mathrm{b}}}{2\mathrm{\pi I}}\mathrm{\pi }\left({\mathrm{s}}^2-{\mathrm{a}}^2\right) \\ \varnothing =\frac{{\mathrm{\mu }}_0\mathrm{Q}}{2\mathrm{\pi I}}\left({\mathrm{\omega }}_{\mathrm{a}}{\mathrm{a}}^2-{\mathrm{\omega }}_{\mathrm{b}}{\mathrm{s}}^2\right) \end{gathered}$$

Substitute the known values in equation (4).

$$\begin{gathered} E=-\frac{1}{2\mathrm{\pi s}}\frac{\left[{\displaystyle \frac{{\mu }_{0}Q}{2I}}\left({\omega }_a{a}^2-{\omega }_b{s}^2\right)\right]}{dt}\overbrace{\varnothing } \\ E=-\frac{1}{2\mathrm{\pi s}}\frac{{\mu }_{0}Q}{2I}\left(a^2\frac{d{\omega }_a}{dt}-s^2\frac{d{\omega }_b}{dt}\right)\overbrace{\varnothing } \end{gathered}$$

For radius a and b, the induces electric field will be,

$E\left(a\right)=-\frac{{\mu }_{0}Qa}{4\mathrm{\pi I}}\left(\frac{d{\omega }_a}{dt}-\frac{d{\omega }_b}{dt}\right)\overbrace{\varnothing }$and$E\left(b\right)=-\frac{{\mu }_{0}Qa}{4\mathrm{\pi I}}\left(a^2\frac{d{\omega }_a}{dt}-b^2\frac{d{\omega }_b}{dt}\right)\overbrace{\varnothing }$

Write the expression for the torque on a shell.

$$\begin{gathered} N=r\times qE \\ N=qsE\overbrace{z} \end{gathered}$$

For radius a:

$$\begin{gathered} N_a=Q_a\left(-\frac{{\mu }_{0}Qa}{4\mathrm{\pi I}}\right)\left(\frac{d{\omega }_a}{dt}-\frac{d{\omega }_b}{dt}\right)\overbrace{z} \\ {L}_a={\int }_0^{\infty }{N}_a=-\frac{{\mu }_0{Q}^2{a}^2}{4\mathrm{\pi I}}\left({\omega }_a-{\omega }_b\right)\overbrace{z} \end{gathered}$$

For radius b:

$$\begin{gathered} N_b=-Q_b\left(-\frac{{\mu }_{0}Q}{4\mathrm{\pi I}b}\right)\left(a^2\frac{d{\omega }_a}{dt}-b^2\frac{d{\omega }_b}{dt}\right)\overbrace{z} \\ {L}_b={\int }_0^{\infty }{N}_b=-\frac{{\mu }_0{Q}^2}{4\mathrm{\pi I}}\left(a^2{\omega }_a-b^2{\omega }_b\right)\overbrace{z} \end{gathered}$$

Hence, the extra angular momentum will be,

$$\begin{gathered} L_{tot}=L_a+L_B \\ {L}_{tot}=\left(-\frac{{\mu }_0{Q}^2{a}^2}{4\mathrm{\pi I}}\left({\omega }_a-{\omega }_b\right)\overbrace{z}\right)+\left(-\frac{{\mu }_0{Q}^2}{4\mathrm{\pi I}}\left(a^2{\omega }_a-b^2{\omega }_b\right)\overbrace{z}\right) \\ {L}_{tot}=-\frac{{\mu }_0{Q}^2{\omega }_b}{4\mathrm{\pi I}}\left(b^2-a^2\right)\overbrace{z} \end{gathered}$$

So, the reduction in the final angular momentum in part (b) is equal to the residual angular momentum in part (a).

Therefore, the resulting extra angular momentum is $L_{tot}=-\frac{{\mu }_0{Q}^2{\omega }_b}{4\mathrm{\pi I}}\left(b^2-a^2\right)\overbrace{z}$.